Solve Over the Interval sin(2x)=( square root of 2)/2 , 0<=x<2pi

$\sin (2 x) = \frac{\sqrt{2}}{2}$ , $0 \leq x < 2 π$

Step 1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$2 x = \arcsin (\frac{\sqrt{2}}{2})$

Step 2

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$2 x = \frac{π}{4}$

Step 3

Divide each term in $2 x = \frac{π}{4}$ by $2$ and simplify.

Divide each term in $2 x = \frac{π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{4}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{4} \cdot \frac{1}{2}$

Multiply $\frac{π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{π}{4}$ by $\frac{1}{2}$ .

$x = \frac{π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$x = \frac{π}{8}$

Step 4

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$2 x = π - \frac{π}{4}$

Step 5

Solve for $x$ .

Simplify.

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 x = π \cdot \frac{4}{4} - \frac{π}{4}$

Combine $π$ and $\frac{4}{4}$ .

$2 x = \frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 4 - π}{4}$

Subtract $π$ from $π \cdot 4$ .

Reorder $π$ and $4$ .

$2 x = \frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 \cdot π$ .

$2 x = \frac{3 \cdot π}{4}$

Divide each term in $2 x = \frac{3 \cdot π}{4}$ by $2$ and simplify.

Divide each term in $2 x = \frac{3 \cdot π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{3 \cdot π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{3 \cdot π}{4}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{3 \cdot π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 \cdot π}{4} \cdot \frac{1}{2}$

Multiply $\frac{3 π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{3 π}{4}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$x = \frac{3 π}{8}$

Step 6

Find the period of $\sin (2 x)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

Step 7

The period of the $\sin (2 x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \frac{π}{8} + π n, \frac{3 π}{8} + π n$ , for any integer $n$

Step 8

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{π}{8} + π (0)$

Simplify.

Multiply $π$ by $0$ .

$\frac{π}{8} + 0$

Add $\frac{π}{8}$ and $0$ .

$\frac{π}{8}$

The interval $[0, 2 π)$ contains $\frac{π}{8}$ .

$x = \frac{π}{8}$

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{3 π}{8} + π (0)$

Simplify.

Multiply $π$ by $0$ .

$\frac{3 π}{8} + 0$

Add $\frac{3 π}{8}$ and $0$ .

$\frac{3 π}{8}$

The interval $[0, 2 π)$ contains $\frac{3 π}{8}$ .

$x = \frac{π}{8}, \frac{3 π}{8}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{π}{8} + π (1)$

Simplify.

Multiply $π$ by $1$ .

$\frac{π}{8} + π$

To write $π$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$\frac{π}{8} + π \cdot \frac{8}{8}$

Combine fractions.

Combine $π$ and $\frac{8}{8}$ .

$\frac{π}{8} + \frac{π \cdot 8}{8}$

Combine the numerators over the common denominator.

$\frac{π + π \cdot 8}{8}$

Simplify the numerator.

Move $8$ to the left of $π$ .

$\frac{π + 8 \cdot π}{8}$

Add $π$ and $8 π$ .

$\frac{9 π}{8}$

The interval $[0, 2 π)$ contains $\frac{9 π}{8}$ .

$x = \frac{π}{8}, \frac{3 π}{8}, \frac{9 π}{8}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{3 π}{8} + π (1)$

Simplify.

Multiply $π$ by $1$ .

$\frac{3 π}{8} + π$

To write $π$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$\frac{3 π}{8} + π \cdot \frac{8}{8}$

Combine fractions.

Combine $π$ and $\frac{8}{8}$ .

$\frac{3 π}{8} + \frac{π \cdot 8}{8}$

Combine the numerators over the common denominator.

$\frac{3 π + π \cdot 8}{8}$

Simplify the numerator.

Move $8$ to the left of $π$ .

$\frac{3 π + 8 \cdot π}{8}$

Add $3 π$ and $8 π$ .

$\frac{11 π}{8}$

The interval $[0, 2 π)$ contains $\frac{11 π}{8}$ .

$x = \frac{π}{8}, \frac{3 π}{8}, \frac{9 π}{8}, \frac{11 π}{8}$

Solve Over the Interval sin(2x)=( square root of 2)/2 , 0<=x<2pi

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